Bounded sequence in $L^1$ with no weakly convergent subsequence and bounded sequence in $L^\infty$ with no weakly convergent subsequence Can you give two elementary examples that show: 


*

*a bounded sequence in $L^1$ with no weakly convergent subsequence;

*a bounded sequence in $L^\infty$ with no weakly convergent subsequence??



For the $L^1$ case, I was thinking about an approximation of the Dirac delta, say rectangles of area 1 that get higher and smaller around the origin, but I cannot see if I can make it work. I'm totally lost about the $L^\infty$ case.
 A: For the case of $L^1$ your example does work. Let $f_n= n\chi_{(0,1/n)}.$ Suppose, to reach a contradiction, that $f_{n_k}$ converges weakly in $L^1.$ Passing to a further subsequence, which I'll still denote by $f_{n_k},$ we can assume $n_{k+1}/n_k \to \infty.$ Define $g\in L^\infty$ by setting $g=(-1)^k$ on $(1/n_k, 1/n_{k+1}).$ Then $\int_0^1 f_{n_k}g \to -1$ as $k\to \infty$ through odd integers, while $\int_0^1 f_{n_k}g \to 1$ through even integers. Thus $\int_0^1 f_{n_k}g$ fails to have a limit as $k\to \infty,$ contradiction.

Added later: Why do we need $n_k/n_{k+1} \to \infty?$ Hopefully this will help:
$$\int_0^1 f_{n_k}g = n_k\int_0^{1/n_k} g
= n_k\sum_{j=k}^{\infty}(-1)^j(1/n_j- 1/n_{j+1}) = (-1)^k(1- n_k/n_{k+1}) + r_k.$$
Verify that $|r_k| \le n_k/n_{k+1}.$ Thus the above integrals have the behavior described.
A: Let me first prove that your example in the $L^1$ case works. Let
$$ f_n=n 1_{(-1/(2n), 1/(2n))}.$$
Assume by contradiction that your sequence has a weakly convergent subsequence and let $f\in L^1$ be the weak limit. Let 
$$M_{+,\epsilon}=\{x\in \mathbb{R}\setminus (-\epsilon, \epsilon): f(x)>0\}$$ 
and
$$M_{-,\epsilon}=\{x\in \mathbb{R}\setminus (-\epsilon, \epsilon): f(x)<0\}.$$
Then we define the functionals $T_{\pm,\epsilon}(g)=\int_{M_{\pm ,\epsilon}} g$. Using weak convergence, one has $T_{\pm , \epsilon}(f)=0$. From this we obtain by the monotone convergence theorem that
$$\Vert f \Vert = \lim_{\epsilon \rightarrow 0} T_{+, \epsilon}(f) - T_{-, \epsilon}(f) =0.$$
This implies $f=0$. Define $T(g)=\int_{-1}^1 g$. For every member of our sequence, this functional equal 1, however, $T(f)=0$ which gives you a contradiction.
For the $L^\infty$ case set
$$f_n=(1-n\vert x \vert )1_{(-1/n,1/n)}$$
Similarly as above, one proves that $f=0$ (if it wasn't zero, there would be $C,R>0$ such that $M_+=\{f\geq C\}\cap (-R, R)$ or $M_-=\{f<-C\}\cap (-R, R)$ is not a null set. Consider then the functionals $L_\pm(g)=\int_{M_\pm} g$. By dominated convergence, we get the contradiction $L_\pm (f)=0$). Now define
$$ T: C^0(\mathbb{R}, \mathbb{R}) \rightarrow \mathbb{R}, \ T(g)=g(0)$$
Using Hahn-Banach, we can extend this functional to all of $L^\infty$. Call the extension $S$. Then we get the contradiction (note that the $f_n$ are continuous and $f_n(0)=1$)
$$ 0=S(f)= \lim_{n\rightarrow \infty} S(f_n)=1.$$
